question_answer

Locus of a point which moves such that its distance from the line \[x-y=0\]is

A) \[{{x}^{2}}+4xy-{{y}^{2}}=0\]

B) \[2{{x}^{2}}-4xy+{{y}^{2}}=0\]

C) \[{{x}^{2}}-4xy+{{y}^{2}}=0\]

D) \[{{x}^{2}}-4xy-{{y}^{2}}=0\]

Correct Answer: B

Solution :

\[{{P}_{1}}=\]length of perpendicular from P to \[x\text{-}\]axis. \[{{P}_{2}}=\]length of perpendicular from P to \[y=x\]line. \[{{P}_{1}}=2{{P}_{2}}\] \[\left| \,k\, \right|=2\cdot \frac{\left| h-k \right|}{\sqrt{2}}\] Squaring on both sides. \[{{k}^{2}}=2\,{{(h-k)}^{2}}\] \[{{k}^{2}}=2{{h}^{2}}+2{{k}^{2}}-4hk\] \[\Rightarrow \]               \[2{{h}^{2}}-4hk+{{k}^{2}}=0\] So, the locus of a point P is, \[2{{x}^{2}}-4xy+{{y}^{2}}=0\]